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\begin{document}

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\begin{center}
\vskip 1cm{\LARGE\bf Sums of Products of  Bernoulli Numbers, \\
\vskip .1in
Including Poly-Bernoulli Numbers
}
\vskip 1cm
\large
Ken Kamano \\
Department of General Education\\
Salesian Polytechnic \\
4-6-8, Oyamagaoka, Machida-city, Tokyo 194-0215\\
Japan\\
\href{mailto:kamano@salesio-sp.ac.jp}{\tt kamano@salesio-sp.ac.jp}
\end{center}

\vskip .2 in

\begin{abstract}
We investigate sums of products of Bernoulli numbers including
poly-Bernoulli numbers.  A relation among these sums and explicit
expressions of sums of two and three products are given.  As a
corollary, we obtain fractional parts of sums of two and three products
for negative indices.
\end{abstract}

\newtheorem{theorem}{Theorem}
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\def\Li{{\rm Li}}



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\section{Introduction and main results}

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Bernoulli numbers $B_n$ ($n=0,1,2,\ldots$) are defined by the following generating function:
\[ \frac{t}{e^t-1} = \sum_{n=0}^{\infty} \frac{B_n}{n!}t^n. \]
The following identity on sums of two products of Bernoulli numbers
is known as Euler's formula:
\begin{equation}\label{eq:Euler1}
 \sum_{i=0}^n \binom{n}{i} B_iB_{n-i} = -nB_{n-1} -(n-1)B_{n} \ \ (n\ge 1).
\end{equation}
When $n$ is an even integer, the identity \eqref{eq:Euler1} can be written as
\begin{equation}\label{eq:Euler2}
 \sum_{i=1}^{n-1} \binom{2n}{2i} B_{2i}B_{2n-2i} = -(2n+1)B_{2n}\ \ (n\ge 2),
\end{equation}
because $B_n=0$ for any odd integer $n\ge 3$.
Many generalizations of \eqref{eq:Euler1} and 
\eqref{eq:Euler2} have been considered.
As a generalization of \eqref{eq:Euler2}, 
Dilcher \cite{D} gave closed formulas of sums of $N$ products of Bernoulli numbers
for any positive integer $N$.
Chen \cite{Chen} gave generalizations of \eqref{eq:Euler1} for 
sums of $N$ products of Bernoulli polynomials, generalized Bernoulli numbers
and Euler polynomials
by using special values of certain zeta functions at non-positive integers.
Other types of sums of products have been also studied;
see, for example, \cite{AD2007,AD2009,E Trans,Ma,Pe,PS}.

The reason why these formulas are valid
is that the generating function of Bernoulli numbers 
satisfies simple differential equations.
For example, Euler's formula \eqref{eq:Euler1} is derived
by comparing the coefficients of the following identity:
\begin{equation}\label{eq:diff.eq.of F_1}
 F(t)^2 = -tF'(t) +(1-t)F(t), \end{equation}
where $F(t)=t/(e^t-1)$.



For any integer $k$, Kaneko \cite{K} 
introduced poly-Bernoulli numbers of index $k$
(denoted by $B_n^{(k)}$) by the following 
generating function:
\begin{equation}\label{eq:poly-Bernoulli defi} \frac{\Li_k(1-e^{-t})}{1-e^{-t}} 
= \sum_{n=0}^{\infty} \frac{B_n^{(k)}}{n!} t^n,
\end{equation}
where $\Li_k(x)$ is the $k$-th polylogarithm defined by
$\Li_k(x) = \sum_{n=1}^{\infty} x^n/n^k$.
The list of poly-Bernoulli numbers $B_n^{(k)}$ with $-5\le k \le 5$ and 
$0\le n \le 7$ are given by Arakawa and Kaneko \cite{AK}.
The numbers $B_n^{(k)}$ are rational numbers, 
in particular, are positive integers for $k\le 0$ (e.g., \cite[Section 1]{K}).
When $k=1$, the left-hand side of \eqref{eq:poly-Bernoulli defi} is 
equal to $te^t/(e^t-1)$ because of $\Li_1(x)=-\log(1-x)$. Since
\begin{equation}\label{eq:alternate Bernoulli}
 \frac{te^t}{e^t-1} = \sum_{n=0}^{\infty} \frac{(-1)^nB_n}{n!} t^n,
\end{equation} 
we have $B_n^{(1)} =(-1)^nB_n$ for $n\ge 0$
(actually $B_n^{(1)}=B_n$ except for $n=1$).
Poly-Bernoulli numbers of positive index are related to multiple zeta functions.
To be more precise,
special values of certain multiple zeta functions
at non-positive integers
are described in terms of 
poly-Bernoulli numbers (cf.\,\cite{AK_Nagoya}).
For combinatorial interpretations of
poly-Bernoulli numbers of negative index,
see Brewbaker \cite{B} and Launois \cite{L}.




In this paper we investigate the following type of sums of products 
of Bernoulli numbers including poly-Bernoulli numbers:
\begin{equation}\label{eq:sop invol. poly}
S_m^{(k)}(n) := 
\sum_{\substack{i_1+\cdots +i_{m}=n \\ i_1,\ldots ,i_{m}\ge 0}}
\binom{n}{i_1,\ldots ,i_{m}}
 B_{i_1}\cdots B_{i_{m-1}} B_{i_{m}}^{(k)}\ \ (m\ge 1, n\ge 0),
\end{equation}
where 
$\binom{n}{i_1,\ldots ,i_{m}}$ are
multinomial coefficients defined by 
\begin{equation*}
 \binom{n}{i_1,\ldots,i_{m}} = \frac{n!}{i_1! \cdots i_m!}.
\end{equation*}
Clearly, it holds that $S_1^{(k)}(n) = B_n^{(k)}$.
We list $S_2^{(k)}(n)$ and $S_3^{(k)}(n)$ with $-4\le k \le 4$ and $0\le n \le 6$
in Tables \ref{tbl001} and \ref{tbl002}.
We note that the numbers $S_m^{(k)}(n)$ appear as the coefficients of the following 
generating function:
\begin{equation}\label{SP-generating}
 \left(\frac{t}{e^t-1}\right)^{m-1} \frac{\Li_k(1-e^{-t})}{1-e^{-t}}
= \sum_{n=0}^{\infty} S_m^{(k)}(n) \frac{t^n}{n!}.
\end{equation}
The left-hand side of \eqref{SP-generating} satisfies 
a certain differential equation like \eqref{eq:diff.eq.of F_1}
(see Proposition \ref{prop:dm_equation}),
thus this type of sums \eqref{eq:sop invol. poly}
is one of natural extensions of the classical sums of products of Bernoulli numbers.


\begin{table}[t]
\caption{$S_2^{(k)}(n)$}\label{tbl001}\vspace{-10pt}
\begin{center}
{\renewcommand\arraystretch{1.3}
$\begin{array}{|c||c|c|c|c|c|c|c|}
\hline
 k \backslash n &amp;0&amp; 1 &amp; 2&amp;3&amp; 4 &amp;5 &amp; 6 \\
\hline
\hline 
-4 &amp; 1 &amp; \frac{31}{2}  &amp;\frac{781}{6}  &amp; 855       &amp;\frac{147479}{30}&amp; 26025 &amp; \frac{5474701}{42} \\
\hline
-3 &amp; 1 &amp; \frac{15}{2} &amp;\frac{229}{6}  &amp; 165        &amp; \frac{19559}{30} &amp; 2435 &amp; \frac{367669}{42}\\
\hline
-2 &amp; 1 &amp; \frac{7}{2} &amp;\frac{61}{6}  &amp;  27          &amp; \frac{2039}{30}  &amp;165 &amp; \frac{16381}{42}   \\
\hline
-1 &amp; 1 &amp; \frac{3}{2} &amp; \frac{13}{6} &amp; 3            &amp; \frac{119}{30} &amp; 5  &amp; \frac{253}{42}    \\
\hline
0 &amp; 1 &amp; \frac{1}{2} &amp;\frac{1}{6}    &amp; 0             &amp; -\frac{1}{30} &amp; 0  &amp; \frac{1}{42}     \\
\hline
1 &amp; 1 &amp; 0           &amp;-\frac{1}{6}   &amp;  0           &amp; \frac{1}{10}  &amp;  0  &amp; -\frac{5}{42}     \\
\hline
2 &amp; 1 &amp; -\frac{1}{4} &amp; -\frac{1}{9} &amp; \frac{1}{8} &amp; \frac{17}{450}  &amp; -\frac{1}{8} &amp; -\frac{23}{1470} \\
\hline
3 &amp; 1 &amp; -\frac{3}{8} &amp; -\frac{1}{108} &amp; \frac{13}{96} &amp; -\frac{733}{13500}  &amp;-\frac{131}{1440} 
 &amp; \frac{65953}{617400}\\
\hline
4 &amp; 1  &amp; -\frac{7}{16} &amp; \frac{43}{648} &amp;  \frac{115}{1152}  &amp; -\frac{70271}{810000} &amp;
  -\frac{233}{9600} &amp; \frac{26855027}{259308000} \\
\hline
\end{array}$}
\end{center}
\end{table}
\begin{table}[t]
\caption{$S_3^{(k)}(n)$}\label{tbl002}\vspace{-10pt}
\begin{center}
{\renewcommand\arraystretch{1.3}
$\begin{array}{|c||c|c|c|c|c|c|c|}
\hline
 k \backslash n &amp;0&amp; 1 &amp; 2&amp;3&amp; 4 &amp;5 &amp; 6 \\
\hline
\hline
-4 &amp; 1 &amp; 15  &amp;\frac{689}{6} &amp;\frac{1335}{2}  &amp; \frac{33361}{10}   &amp; \frac{30315}{2}  &amp; \frac{2708995}{42}\\
\hline
-3 &amp; 1 &amp; 7 &amp;\frac{185}{6}  &amp; \frac{223}{2} &amp; \frac{3601}{10} &amp; \frac{6473}{6}  &amp; \frac{128515}{42}\\
\hline
-2 &amp; 1 &amp; 3 &amp;\frac{41}{6}  &amp; \frac{27}{2} &amp; \frac{241}{10}  &amp; \frac{79}{2}  &amp; \frac{2515}{42}\\
\hline
-1 &amp; 1 &amp; 1 &amp; \frac{5}{6} &amp; \frac{1}{2} &amp; \frac{1}{10} &amp; -\frac{1}{6}   &amp; -\frac{5}{42}\\
\hline
0 &amp; 1 &amp; 0 &amp;-\frac{1}{6}  &amp; 0 &amp; \frac{1}{10} &amp; 0  &amp; -\frac{5}{42}\\
\hline
1 &amp; 1 &amp; -\frac{1}{2} &amp; 0 &amp; \frac{1}{4} &amp; -\frac{1}{10}  &amp; -\frac{1}{4}  &amp; \frac{5}{21}\\
\hline
2 &amp; 1 &amp; -\frac{3}{4} &amp; \frac{11}{36} &amp; \frac{1}{6} &amp; -\frac{107}{300}  &amp; \frac{11}{360}  &amp; \frac{209}{392}\\
\hline
3 &amp; 1 &amp; -\frac{7}{8} &amp; \frac{115}{216} &amp; -\frac{11}{288} &amp; -\frac{6619}{18000}  &amp; \frac{899}{2700} 
 &amp; \frac{134563}{493920}\\
\hline
4 &amp; 1 &amp; -\frac{15}{16}  &amp;\frac{869}{1296}  &amp;-\frac{755}{3456}&amp;-\frac{273653}{1080000}&amp; \frac{279877}{648000} 
 &amp; -\frac{10347133}{207446400}\\
\hline
\end{array}$}
\end{center}
\end{table}


Now we state our main results of this paper.
\begin{thm}\label{thm:maintheorem}
For  $k\in\Z$ and $m\ge 1$, we have
\begin{equation}\label{eq:maintheorem}
\begin{split}
&amp;\sum_{l=0}^{m}
(-1)^{m-l}  {m+1 \atopwithdelims[] l+1} 
S_{m+1}^{(k-l)}(n)\\
&amp;=
\begin{cases}
 n(n-1)\cdots (n-m+1) 
\displaystyle \sum_{l=1}^m 
{m \atopwithdelims[] l} B_{n-m+l}^{(k)} \, ,  &amp; \text{\ \  if } n\ge m;\\
0 , &amp; \text{\ \  if } 0\le n \le m-1,
\end{cases}
\end{split}
\end{equation}
where 
${m \atopwithdelims[] l}$ are (unsigned) Stirling numbers of the first kind.
\end{thm}


The definition of Stirling numbers of the first kind ${m \atopwithdelims[] l}$
will be given in Section 2.
Although Theorem \ref{thm:maintheorem} only gives relations among sums of products $S_m^{(k)}(n)$, 
explicit formulas of $S_m^{(k)}(n)$ can be obtained for $m=2$ and $3$.

\begin{thm}\label{thm:m=1 theorem}
For $k\ge 1$ and $n\ge 0$, it holds that
\begin{align}
S_2^{(0)}(n)
&amp;= B_n^{(1)}, \label{eq:m=1theorem_0}\\
S_2^{(k)}(n)
&amp;= B_n^{(1)} -n \sum_{j=1}^k B_n^{(j)}, \label{eq:m=1theorem}\\
S_2^{(-k)}(n)
&amp;= B_n^{(1)} +n \sum_{j=0}^{k-1} B_n^{(-j)}. \label{eq:m=1theorem_negative}
\end{align}
\end{thm}
When $k=1$ in \eqref{eq:m=1theorem}, 
we have 
\begin{equation}\label{eq:cor transform 1 to -1}
 \sum_{i=0}^n \binom{n}{i}(-1)^{n-i} B_{i} B_{n-i}
= -(n-1)B_{n}.
\end{equation} 
The identity \eqref{eq:cor transform 1 to -1} is equivalent to \eqref{eq:Euler1} because 
$B_n=0$ for any odd integer $n\ge 3$.
Therefore Theorem \ref{thm:m=1 theorem} can be regarded as a generalization of 
Euler's formula \eqref{eq:Euler1}.


\begin{thm}\label{thm:m=2 theorem}
For $k\ge 1$ and $n\ge 1$, it holds that
\begin{align}
S_3^{(0)}(n) =&amp; -(n-1)B_n, \label{eq:m=2theorem_0} \\
\begin{split}
 S_3^{(k)}(n)
=&amp;
(-1)^n ( 1-2^{-k}) B_{n-1}
-(n-1)B_n\\
&amp;+n(n-1) \sum_{j=1}^k (1-2^{j-k-1}) \left(B_n^{(j)}+B_{n-1}^{(j)}\right),
\end{split} \label{eq:m=2 theorem} \\
\begin{split}
S_3^{(-k)}(n) 
=&amp;
n  (2^k-1)(-1)^{n-1} B_{n-1} -(n-1)B_n \\
 &amp; +n(n-1) \sum_{j=0}^{k-2} (2^{k-1-j}-1) \left(B_n^{(-j)} + B_{n-1}^{(-j)}\right).
\end{split} \label{eq:m=2 theorem negative}
\end{align}
\end{thm}
As a corollary,
for a negative index $-k$, we obtain the following formulas on
fractional parts of $S_2^{(-k)}(n)$ and $S_3^{(-k)}(n)$.
\begin{cor}\label{cor:denominator}
For $k\ge 1$ and $n\ge 0$, we have
\begin{align}
 S_2^{(-k)}(n)  &amp;\equiv   B_n \pmod{1},   \label{eq:denominator S_2}\\
 S_3^{(-k)}(n) &amp;\equiv 
\begin{cases}
n(2^k-1)B_{n-1} \pmod{1}, &amp; $if $ n $ is odd$; \\
-(n-1)B_{n} \pmod{1}, &amp; $if $ n $ is even$.
\end{cases} \label{eq:denominator S_3}
\end{align}
Here $\alpha \equiv \beta \pmod{1}$ means
$\alpha - \beta \in\Z $ for rational numbers $\alpha$ and $\beta$.
\end{cor}


The classical von Staudt-Clausen theorem (e.g., \cite[Section 7.9]{H-W}) states that
\begin{equation*}
 B_n \equiv -\sum_{\substack{p : \text{prime} \\ (p-1)\mid n}} \frac{1}{p} \pmod{1}
\end{equation*}
for any even integer $n\ge 2$. Therefore, by using Corollary \ref{cor:denominator},
we can determine the fractional parts of $S_2^{(-k)}(n)$ and $S_3^{(-k)}(n)$
if $k\ge 1$ and $n\ge 0$ are given.


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\section{Proof of Theorem \ref{thm:maintheorem}}

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We first recall (unsigned) Stirling numbers of the first kind.
Let $m$ be a positive integer. 
For $0\le l\le m$, Stirling numbers of the first kind
${m \atopwithdelims[] l}$ are defined as
\begin{equation}
 x(x+1)\cdots (x+m-1) = \sum_{l=0}^m{m \atopwithdelims[] l} x^l.
\end{equation}
It follows immediately that ${m \atopwithdelims[] 0}=0$
and ${m \atopwithdelims[] m} = 1$ for all $m\ge 1$.
For $l\ge m+1$ and $l\le -1$, we define ${m \atopwithdelims[] l}=0$.
Then the recurrence relation
\begin{equation}\label{eq:Stirling recurrence}
{m+1 \atopwithdelims[] l}
 = {m \atopwithdelims[] l-1} + m{m \atopwithdelims[] l}\end{equation}
holds for all $m\ge 1$ and $l\in\Z$.

We set the generating function of poly-Bernoulli numbers of index $k$ as 
$F_k(t)$, 
i.e.,
\[ F_k(t) := \frac{\Li_k(1-e^{-t})}{1-e^{-t}}.\]
For $k=1$, $0$ and $-1$, they have simple expressions;
$ F_1(t)=te^t/(e^t-1)$, $F_0(t) = e^t$ and $F_{-1}(t)= e^{2t}$. 

Now let us prove Theorem \ref{thm:maintheorem}.
The $n$-th coefficient of $t^m \frac{d^l}{dt^l}F_k(t)$ is equal to 
\[ 
\begin{cases}
 \dfrac{n(n-1)\cdots (n-m+1) B_{n-m+l}^{(k)}}{n!},  &amp; \text{ if } n\ge m; \\
 0 , &amp; \text{ if } 0\le n\le m-1.
\end{cases} \]
Therefore it suffices to show the following proposition
to get Theorem \ref{thm:maintheorem}.

\begin{prop}\label{prop:dm_equation}
For $k\in \Z$ and $m\ge 1$ we have
\begin{equation}\label{eq:dm_equation}
\begin{split}
&amp; \left(
 {m \atopwithdelims[] m} \frac{d^{m}}{dt^m} + 
 {m \atopwithdelims[] m-1} \frac{d^{m-1}}{dt^{m-1}} + 
 \cdots +
 {m \atopwithdelims[] 1} \frac{d}{dt}
  \right) F_k(t) \\
&amp;=
\frac{1}{(e^t-1)^m}
\sum_{l=0}^{m}
(-1)^{m-l}  {m+1 \atopwithdelims[] l+1} F_{k-l} (t).
\end{split}
\end{equation}
\end{prop}
\vspace{10pt}


\begin{proof} 
We prove the proposition by induction on $m$.
Since $\frac{d}{dt} F_k(t) = F_{k-1}(t)/t$, 
we can easily prove that
\begin{equation}\label{eq:diffeq_F_k}
\frac{d}{dt}F_k(t)= \frac{1}{e^t-1} (F_{k-1}(t) - F_k(t))\ \ \ \ (k\in\Z).
\end{equation}
Hence the case $m=1$ holds.

We assume that \eqref{eq:dm_equation} holds for a certain $m$.
By \eqref{eq:Stirling recurrence}, we have
\begin{equation}\label{eq:prop1}
\begin{split}
&amp; \left(
 {m+1 \atopwithdelims[] m+1} \frac{d^{m+1}}{dt^{m+1}} + 
 {m+1 \atopwithdelims[] m} \frac{d^{m}}{dt^{m}} + 
 \cdots +
 {m+1 \atopwithdelims[] 1} \frac{d}{dt}
  \right) F_k(t) \\
&amp; = \frac{d}{dt} \left(
 {m \atopwithdelims[] m} \frac{d^{m}}{dt^{m}} + 
\cdots +
 {m \atopwithdelims[] 1} \frac{d}{dt}
  \right) F_k(t) +m \left(
 {m \atopwithdelims[] m} \frac{d^{m}}{dt^{m}} + 
\cdots +
 {m \atopwithdelims[] 1} \frac{d}{dt}
  \right) F_k(t).
\end{split}
\end{equation}
By the inductive assumption and \eqref{eq:diffeq_F_k}, the right-hand side of \eqref{eq:prop1} is equal to
\begin{align*}
&amp; \frac{d}{dt}
 \left(
 \frac{1}{(e^t-1)^m}
 \sum_{l=0}^{m}
 (-1)^{m-l}  {m+1 \atopwithdelims[] l+1} F_{k-l} (t)
 \right)
 +
 \frac{m}{(e^t-1)^m}
 \sum_{l=0}^{m}
 (-1)^{m-l}  {m+1 \atopwithdelims[] l+1} F_{k-l} (t) \\
=&amp; 
 \frac{-me^t}{(e^t-1)^{m+1}}
 \sum_{l=0}^{m}
 (-1)^{m-l}  {m+1 \atopwithdelims[] l+1} F_{k-l} (t)\\
&amp;+ 
 \frac{1}{(e^t-1)^{m+1}}
 \sum_{l=0}^{m}
 (-1)^{m-l}  {m+1 \atopwithdelims[] l+1} (F_{k-l-1} (t) - F_{k-l}(t)) \\
&amp;+
 \frac{m}{(e^t-1)^m}
 \sum_{l=0}^{m}
 (-1)^{m-l}  {m+1 \atopwithdelims[] l+1} F_{k-l} (t) \\
=&amp; \frac{-m-1}{(e^t-1)^{m+1}}
 \sum_{l=0}^{m}
 (-1)^{m-l}  {m+1 \atopwithdelims[] l+1} F_{k-l} (t)\\
&amp;+ 
 \frac{1}{(e^t-1)^{m+1}}
 \sum_{l=0}^{m}
 (-1)^{m-l}  {m+1 \atopwithdelims[] l+1} F_{k-l-1} (t)  \\
=&amp; \frac{1}{(e^t-1)^{m+1}}
 \sum_{l=0}^{m+1}
 (-1)^{(m+1)-l} 
  \left( (m+1){m+1 \atopwithdelims[] l+1} + {m+1 \atopwithdelims[] l} \right)
    F_{k-l} (t).
\end{align*}
As a consequence, by using the relation \eqref{eq:Stirling recurrence} again, 
we obtain
\begin{align*}
&amp; \left(
 {m+1 \atopwithdelims[] m+1} \frac{d^{m+1}}{dt^{m+1}} + 
 {m+1 \atopwithdelims[] m} \frac{d^{m}}{dt^{m}} + 
 \cdots +
 {m+1 \atopwithdelims[] 1} \frac{d}{dt}
  \right) F_k(t) \\
=&amp; \frac{1}{(e^t-1)^{m+1}}
 \sum_{l=0}^{m+1}
 (-1)^{(m+1)-l} {(m+1)+1 \atopwithdelims[] l+1} 
    F_{k-l} (t).
\end{align*}
Therefore \eqref{eq:dm_equation} also holds for $m+1$
and this completes the proof.
\end{proof}





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\section{Explicit formulas of $\boldsymbol{S_2^{(k)}(n)}$ and $\boldsymbol{S_3^{(k)}(n)}$}

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In this section, we prove Theorem \ref{thm:m=1 theorem},
Theorem \ref{thm:m=2 theorem}
and Corollary \ref{cor:denominator}.

\begin{proof}[Proof of Theorem \ref{thm:m=1 theorem}]
First we prove \eqref{eq:m=1theorem_0}.
We recall $F_0(t)=e^t$.
By setting $m=2$ and $k=0$ in \eqref{SP-generating},
we obtain that the generating function of $S_2^{(0)}(n)$
is equal to $te^t/(e^t-1)$. Then \eqref{eq:m=1theorem_0} follows from
\eqref{eq:alternate Bernoulli}.

Next we prove the positive index case \eqref{eq:m=1theorem}.
Since the negative index case \eqref{eq:m=1theorem_negative} can be proved similarly,
we omit its proof.
By \eqref{eq:diffeq_F_k}, we have
\[ \sum_{j=1}^k  \frac{d}{dt} F_j(t) = 
\frac{1}{e^t-1} (F_{0}(t) -F_k(t) ). \]
Since $F_0(t)=e^t$, it holds that
\[\frac{t}{e^t-1}F_k(t) = \frac{te^t}{e^t-1} -t \sum_{j=1}^k \frac{d}{dt} F_j(t).\]
By comparing the coefficients of both sides,
we obtain \eqref{eq:m=1theorem}.
\end{proof}


\begin{proof}[Proof of Theorem \ref{thm:m=2 theorem}]
By setting $m=3$ and $k=0$ in \eqref{SP-generating},
we obtain that the generating function of $S_3^{(0)}(n)$ is 
$t^2e^t/(e^t-1)^2$.
This is exactly the same as the generating function of $S_2^{(1)}(n)$,
therefore \eqref{eq:m=2theorem_0} follows from the relation \eqref{eq:cor transform 1 to -1}.

We prove the positive index case \eqref{eq:m=2 theorem}.
We also omit the proof of the negative index case \eqref{eq:m=2 theorem negative}
because it can be proved similarly.
Setting $m=2$ in Proposition \ref{prop:dm_equation}, we have
\begin{equation}\label{eq:pf_thm:m=2_first}
\left(\frac{d^2}{dt^2} +\frac{d}{dt}\right)F_k(t)=
 \frac{1}{(e^t-1)^2} \left((2F_k(t)-F_{k-1}(t))- (2F_{k-1}(t)-F_{k-2}(t))\right).
\end{equation}
By this equation, we get
\begin{equation}\label{eq:pf_thm:m=2_second}
 \frac{2F_l(t)}{(e^t-1)^2} - \frac{F_{l-1}(t)}{(e^t-1)^2} = 
 \frac{2F_0(t)-F_{-1}(t)}{(e^t-1)^2} 
+\sum_{j=1}^l \left(\frac{d^2}{dt^2} +\frac{d}{dt}\right)F_j(t). 
\end{equation}
In fact, this can be proved by replacing $k$ with $j$ in \eqref{eq:pf_thm:m=2_first} and 
summing over $j$ from $1$ to $l$.
Furthermore we multiply both sides of \eqref{eq:pf_thm:m=2_second} by $2^{l-1}$ and
sum over $l$ from $1$ to $k$.
Then we obtain
\begin{align*}
2^k \frac{F_k(t)}{(e^t-1)^2}
=&amp; \frac{F_0(t)}{(e^t-1)^2} + \left(\sum_{l=1}^k 2^{l-1}\right) \frac{2F_0(t)-F_{-1}(t)}{(e^t-1)^2}\\
 &amp;  + \sum_{l=1}^k 2^{l-1} \sum_{j=1}^l \left(\frac{d^2}{dt^2} +\frac{d}{dt}\right)F_j(t) \\
=&amp; \frac{(2^{k+1}-1)F_0(t) - (2^k-1)F_{-1}(t)}{(e^t-1)^2}\\
 &amp; +  \sum_{j=1}^k \left(\sum_{l=j}^ k 2^{l-1}\right) \left(\frac{d^2}{dt^2} +\frac{d}{dt}\right)F_j(t).
\end{align*}
Hence we have
\begin{equation}
\begin{split}
2^k \left(\frac{t}{e^t-1}\right)^2 F_k(t)
=&amp; (2^{k+1}-1) \frac{t}{e^t-1}  \frac{te^t}{e^t-1}
  - (2^k-1)\frac{te^t}{e^t-1}\frac{te^t}{e^t-1}\\
 &amp; + t^2 \sum_{j=1}^k  (2^k-2^{j-1})
  \left(\frac{d^2}{dt^2} +\frac{d}{dt}\right)F_j(t).
\end{split}
\end{equation} 
By comparing the coefficients of both sides, 
we obtain for $n\ge 1$
\begin{equation*}
\begin{split}
2^k S_3^{(k)}(n)
=&amp; (2^{k+1}-1) \sum_{i=0}^n
 \binom{n}{i} (-1)^{n-i} B_{i} B_{n-i} 
  - (2^k-1)\sum_{i=0}^n \binom{n}{i} (-1)^{n} B_{i} B_{n-i} \\
 &amp; +n(n-1) \sum_{j=1}^k  (2^k-2^{j-1})
  \left(B_n^{(j)} + B_{n-1}^{(j)} \right).
\end{split}
\end{equation*}
By \eqref{eq:Euler1}, \eqref{eq:cor transform 1 to -1} and the fact
$(-1)^n(n-1)B_n = (n-1)B_n$ for all $n\ge 1$, it holds that
\begin{equation*}
\begin{split}
&amp; (2^{k+1}-1) \sum_{i=0}^n
 \binom{n}{i} (-1)^{n-i} B_{i} B_{n-i} 
  - (2^k-1)\sum_{i=0}^n \binom{n}{i} (-1)^{n} B_{i} B_{n-i} \\
&amp;= -(2^{k+1}-1)(n-1)B_n - (-1)^n(2^k-1) (-nB_{n-1} - (n-1)B_n)\\
&amp;=-n(2^k-1) (-1)^{n-1} B_{n-1} -(n-1)2^kB_n.
\end{split}
\end{equation*}
Therefore we obtain 
\begin{equation}\label{eq: 2^k}
\begin{split}
2^k S_3^{(k)}(n)
=&amp;
-n\left(2^k- 1\right) (-1)^{n-1} B_{n-1}
-(n-1)2^k B_n \\
&amp; +n(n-1) \sum_{j=1}^k (2^k -2^{j-1}) (B_n^{(j)}+B_{n-1}^{(j)}).
\end{split}
\end{equation} 
Dividing both sides of \eqref{eq: 2^k} by $2^k$, we get 
\eqref{eq:m=2 theorem}.
\end{proof}



\begin{proof}[Proof of Corollary \ref{cor:denominator}]
The congruence \eqref{eq:denominator S_2} immediately follows from 
\eqref{eq:m=1theorem_negative} and
the fact $B_n^{(-k)}$ are integers for $k\ge 0$.

The congruence \eqref{eq:denominator S_3} holds for $n=0$
because $S_3^{(-k)}(0)=1$ for any $k\ge 1$.
We assume that $n\ge 1$.
By \eqref{eq:m=2 theorem negative},
the fractional part of $S_3^{(-k)}(n)$ is 
\begin{equation}\label{eq:fractionalpart} n(2^k-1)(-1)^{n-1}B_{n-1} -(n-1)B_n.
\end{equation}
If $n$ is odd, then $(-1)^{n-1}B_{n-1}=B_{n-1}$ and $(n-1)B_n =0$.
Thus we have $S_3^{(-k)}(n) \equiv n(2^k-1)B_{n-1}$ (mod $1$).
If $n \ge 4$ is even, then $B_{n-1}=0$. 
Thus we have $S_3^{(-k)}(n) \equiv -(n-1)B_{n}$ (mod$1$) for even $n\ge 4$.
This congruence also holds for $n=2$
because the first term of \eqref {eq:fractionalpart} becomes $2^k-1 \in \Z$,
and this completes the proof of \eqref{eq:denominator S_3}.
\end{proof}

\begin{remark}
For $m\ge 4$ we may give explicit formulas of $S_m^{(k)}(n)$
by the method similar to the proof of Theorem \ref{thm:m=1 theorem} and 
Theorem \ref{thm:m=2 theorem}.
However, these formulas seem to be complicated to describe.
\end{remark}








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\bigskip
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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B68; Secondary 11B73.

\noindent \emph{Keywords: } poly-Bernoulli numbers, sums of products.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A027649},
\seqnum{A027650}, and
\seqnum{A027651}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received September 27 2009;
revised version received  April 17 2010.
Published in {\it Journal of Integer Sequences},
April 17 2010.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


\end{document}

                                                                                


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